Conjecture: Let $f:{\mathbb C}^n\rightarrow{\mathbb C}$ be an entire function in $n$ complex variables. Assume that for every $x\in{\mathbb R}^n$ there exists a $y_x\in{\mathbb R}^n$ such that $f(x+iy_x)\equiv f(x_1+iy_{x,1},\ldots,x_n+iy_{x,n})=0$. Then $f\equiv 0$. (We don't assume continuity of $x\mapsto y_x$.)


  1. The conjecture is true for $n=1$: If $f$ satisfies the hypothesis, it has uncountably many different zeros, namely $\{ x+iy_x\ | \ x\in{\mathbb R}\}$. Since ${\mathbb C}$ is $\sigma$-compact, there exists a compact $K\subset{\mathbb C}$ containing uncountably many zeros of $f$. Thus the set of zeros of $f$ has an accumulation point in $K$, and by a well-known result $f$ vanishes identically.

  2. The argument for $n=1$ does not adapt trivially to higher $n$ since holomorphic functions of severable variables have no isolated zeros.

  3. In the conjecture it is important that $x$ consists of the real and $y$ of the imaginary parts of the variables. For example, the following is not true: Assuming that
    $f:{\mathbb C}^2\rightarrow{\mathbb C}$ is entire and for every $z_1=x_1+iy_1$ there is a $z_2=x_2+iy_2$ such that $f(z_1,z_2)=0$, we have $f\equiv 0$. A trivial counterexample is $f(z_1,z_2)=z_1-z_2$.

  4. Most books on holomorphic functions in several variables devote some attention to the set of zeros of such a function. Usually this leads to a proof of Weierstrass' preparation theorem, where the discussion ends. This is of no immediate help to me, since I would actually be happy with a proof of the conjecture for polynomials in $n$ complex variables! In this case, Weierstrass' theorem does not give any new information.


2 Answers 2


The conjecture is obviously false even for $n=2$. Check $f(z,w)=(w-z^2)(z-(w+1)^2)$. Write $z=x+iy$ and $w=u+iv$. Given $x$ and $u$, I can make first term zero unless $u>x^2$. I can make the second term zero unless $x>(u+1)^2$. Since both inequalities can not be true, we are done.

  • $\begingroup$ Thanks. Nice answer. I'll have to see whether there are additional hypotheses in the application that I have in mind. $\endgroup$
    – M Mueger
    Commented Feb 2, 2018 at 17:01

Not a complete answer, but perhaps in the many-variable case you could leverage Theorem 5.1 from

Chirka, E.M., Complex analytic sets. (Kompleksnye analiticheskie mnozhestva), Moskva: ”Nauka”. Glavnaya Redaktsiya Fiziko-Matematicheskoj Literatury. 272 p. R. 3.70 (1985). ZBL0586.32013.

Which states that for an analytic set $Z \subset \mathbb{C}^n$, like the zero set of $f(z) = 0$, the regular part $\operatorname{reg} Z \subset Z$ (which consists of all points around which $Z$ is a complex submanifold) consists of a union of connected components $\operatorname{reg} Z = \bigcup_j S_j$, which is locally finite. That is, for any compact $K\subset \mathbb{C}^n$, $K$ intersects at most finitely many of the $S_j$. The singular part $\operatorname{sng} Z = Z \setminus \operatorname{reg} Z$ is also an analytic set, but of lower dimension. So a similar argument applies to it.

The above reference also contains a wealth of other information about complex analytic sets that might be of use.

  • 1
    $\begingroup$ Well, the zero-set of a holomorphic function is an analytic set and therefore locally a unit of holomorphic (n-1)-manifolds. But I don't see how this helps approaching the conjecture, where real and imaginary parts of the variables are treated differently. $\endgroup$
    – M Mueger
    Commented Feb 2, 2018 at 15:32
  • $\begingroup$ As the counter example in Oleg's answer showed, the obstruction to the conjecture is local (unlike in the $n=1$ case). Had there been non local obstruction, the local finiteness of the union of connected components might have come in useful. $\endgroup$ Commented Feb 3, 2018 at 10:12

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